I also tried resetPage, it sort of works. Here’s what I did:
navigator.push(pageA1);
navigator.push(pageB1);
navigator.push(pageC1);
navigator.resetPage(pageA2); //A2 is component A with a different key.

There are two major constraints with this method:

the reseted page requires a unique key.

onShow on pageA1 will be called (which I don’t really understand why). I ended up having onInit of A2 and onShow of A1 being called at the same time.

What I really want to is to pop back to A1. I am surprised that there is no easy way to deal with this use case.

The Onsen UI tutorial for React regarding the navigation via a side menu does no cover handling the click on a menu item. I do have a question about this part.

Let’s say that I have an app with a side-menu. It is a tree structure; there is no shortcut from a lower level menu item (e.g. Item 1.1) to another lower level item (e.g. Item 2.1) from a different Menu.

I am not making use of Redux by the way. In the App component I initialise the routing for the Root component. I am wondering which construct I should use to render the appropriate page given a click on a menu item (e.g. Menu item 2). Do you have any suggestions for me?

This took me way too long to figure out, and was helped by the people here https://github.com/OnsenUI/OnsenUI/issues/1547
Since Navigator takes an arbitrary route object, basically you pass the props you want to reach your page into the route object. But what if you determine the props at runtime? Pretty easy.